Question

find the positive soltuion​

Answer 1

Step-by-step explanation:

2/(x+1) + 5/2x = 2

2X × 2 + 5(x+1) / (x+1) (x+1) =2

4x+5x+5 /2x^2+2x=2

9x+5=4x^4x

4x^2+4x-9x-5=0

4x^2-5x-5=0

use quadratic formula

x=-b+√b^2-4ac / 2a

x=5+√(-5) ^2-4×4×-5 / 8

x=5+√25+80 / 8

x=5+√105 / 8

Answer is x=5-√105 / 8

Answer 2

Step-by-step explanation:

Given:-

1/(x^2-10x-29) + 1/(x^2-10x-45) - 2/(x^2-10x-69) = 0

To find:-

Find the positive solution ?

Solution:-

Given equation is :

1/(x^2-10x-29) + 1/(x^2-10x-45) - 2/(x^2-10x-69) = 0

It can be written as

1/(x^2-10x-29) + 1/(x^2-10x-29-16) -2(x^2-10x-29-40)

= 0

Put x^2 -10x -29 = a then

=>1/a + 1/(a-16) - 2/(a-40) = 0

=>[(a-16)(a-40)+(a)(a-40)-2(a)(a-16)]/[(a)(a-16)(a-40)]

=0

=>[(a-16)(a-40)+(a)(a-40)-2(a)(a-16)] =

[(a)(a-16)(a-40)]×0

=>[(a-16)(a-40)+(a)(a-40)-2(a)(a-16)] = 0

=>[(a^2-16a-40a+640)+(a^2-40a)-2(a^2-16a)] = 0

=>[a^2-56a+640+a^2-40a-2a^2+32a ] = 0

=>[2a^2-96a+640-2a^2+32a] = 0

=>[-96a+32a+640] = 0

=>[-64a+640] = 0

=>64(-a+10)=0

=> -a+10 = 0/64

=> -a +10 = 0

=> -a = -10

=> a = 10

Now

=>x^2 -10x -29 = 10

=>x^2-10x-29-10 = 0

=>x^2-10x-39 = 0

=>x^2+3x-13x -39 = 0

=>x(x+3)-13(x+3) = 0

=>(x+3)(x-13) = 0

=>x+3 = 0 or x-13 = 0

=>x= -3 or x= 13

The values of x = -3 and 13

The positive value of x = 13

Answer:-

The positive solution for the given equation is 13